Integration:
Integration by Parts
When to use Integration by Parts
Integration by parts is used to evaluate integrals when the usual integration techniques such as substitution and partial fractions can not be applied. The idea of integration by parts is to transform an
integral which cannot be evaluated to an easier integral which can then be evaluated using
one of the
R
0
other integration techniques.
The
formula
that
transforms
the
integral
we
start
with
u(x)v
(x) dx
R 0
into the new one
u (x)v(x) dx is given by:
Z
Z
0
u(x)v (x) dx = u(x)v(x) − u0 (x)v(x) dx
where u(x) and v(x) are two functions of x and u0 (x) =
d
(u(x)),
dx
v 0 (x) =
d
(v(x)).
dx
Choosing u(x) and v 0(x)
To choose which part of our integral function should be u(x) and which should be v 0 (x) we keep the
following in mind:
1. We must be able to differentiate u(x) to get u0 (x) on the RHS.
2. We must be able to integrate v 0 (x) to get v(x) on the RHS.
R 0
R
3. The new integral
u (x)v(x) dx on the RHS should be easier than the original
u(x)v 0 (x) dx
on the LHS.
Keeping the above criteria in mind a general rule of thumb is to choose u(x) to be the part of the
integral expression which is first in the following list which has the mnemonic LIATE:
•
•
•
•
•
Logarithm functions
Inverse Trigonometric functions (Examples: tan−1 (x), cos−1 (x))
Algebraic expressions including polynomials (Examples: x2 , x4 , Constants)
Trigonometric functions (Examples: cos(x), sin(x))
Exponential functions
Note: v 0 (x) should then be chosen to be the part of the expression to be integrated that remains after
the choice of u(x).
Z
Basic Example:
xe2x dx
Using our selection criteria for u(x) we see that that u(x) = x will be our choice as algebraic
expressions (x is a polynomial) comes before exponentials on the LIATE selection list.
u(x) = x
d
(x) = 1
u0 (x) =
dx
v 0 (x) = e2x
Z
e2x
v(x) = e2x dx =
2
Now substituting this into the formula for integration by parts:
Z
Z
0
u(x)v (x) dx = u(x)v(x) − u0 (x)v(x) dx
2x 2x Z
Z
Z
e
xe2x 1
e
2x
− (1)
dx =
−
(x)(e ) dx = (x)
e2x dx
2
2
2
2
Z
xe2x 1 e2x
xe2x e2x
2x
xe dx =
−
+C =
−
+C
2
2 2
2
4
Z
Difficult Example:
cos(2x) ex dx
Using our selection criteria above for u(x) we see that that u(x) = cos(2x) will be our choice as
Trigonometric functions comes before exponentials on the selection list.
u(x) = cos(2x)
d
u0 (x) =
(cos(2x)) = −2 sin(2x)
dx
v 0 (x) = ex
Z
v(x) = ex dx = ex
Now substituting this into the formula for integration by parts:
Z
Z
0
u(x)v (x) dx = u(x)v(x) − u0 (x)v(x) dx
Z
Z
x
x
(cos(2x))(e ) dx = (cos(2x))(e ) − (−2 sin(2x)) (ex ) dx
Z
Z
x
x
cos(2x)e dx = cos(2x)e + 2 sin(2x)ex dx
R
But now it appears we have a problem. The new integral ( sin(2x)ex dx) does not seem any easier
than the original. However if we note the change from the original integral to this new one we can see
if we repeat this integration by parts on the new integral we will arrive back once againRto our starting
integral. We therefore employ integration by parts a second time on the new integral ( sin(2x)ex dx)
where our selection criteria gives us:
u(x) = sin(2x)
d
u0 (x) =
(sin(2x)) = 2 cos(2x)
dx
v 0 (x) = ex
Z
v(x) = ex dx = ex
Now substituting this into the formula for integration by parts:
Z
Z
0
u(x)v (x) dx = u(x)v(x) − u0 (x)v(x) dx
Z
Z
Z
x
x
x
x
(sin(2x))(e ) dx = (sin(2x))(e ) − (2 cos(2x)) (e ) dx = sin(2x)e − 2 cos(2x)ex dx
R
Substituting into the first integration by parts and moving all ( cos(2x)ex dx) integrals to LHS gives:
Z
Z
x
x
cos(2x)e dx = cos(2x)e + 2 sin(2x)ex dx
Z
Z
x
x
x
x
cos(2x)e dx = cos(2x)e + 2 sin(2x)e − 2 cos(2x)e dx
Z
Z
Z
x
x
cos(2x)e dx + 4 cos(2x)e dx = 5 cos(2x)ex dx = cos(2x)ex + 2 sin(2x)ex
Z
1
cos(2x)ex dx = (cos(2x)ex + 2 sin(2x)ex )
5